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Vaselesa [24]
3 years ago
5

Which of the following shows the valency of 1?

Chemistry
2 answers:
Oliga [24]3 years ago
6 0

\huge \bf༆ Answer ༄

The element having valency of 1 is ~

  • Sodium (Na)
Alla [95]3 years ago
6 0

Answer: Pretty sure this is d.) sodium

Explanation:

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gold has a density of 19.32 g/cm3 if you cut a piece of gold in half how much will the density of each piece be?
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Answer:

19.32 g/cm³

Explanation:

The density will remain the same no matter how many times you cut the gold.  The density is g/cm³ or g/mL.  Density is essentially how many grams 1 mL of a compound weighs.

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The pressure exerted by a gas container depends on
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Consider the following reaction at 298.15 K: Co(s)+Fe2+(aq,1.47 M)⟶Co2+(aq,0.33 M)+Fe(s) If the standard reduction potential for
fgiga [73]

Answer:

The correct answer is 0.186 V

Explanation:

The two hemirreactions are:

Reduction: Fe²⁺ + 2 e- → Fe(s)  

Oxidation : Co(s)  → Co²⁺ + 2 e-

Thus, we calculate the standard cell potential (Eº) from the difference between the reduction potentials of cobalt and iron, respectively,  as follows:

Eº = Eº(Fe²⁺/Fe(s)) - Eº(Co²⁺/Co(s)) = -0.28 V - (-0.447 V) = 0.167 V

Then, we use the Nernst equation to calculate the cell potential (E) at 298.15 K:

E= Eº - (0.0592 V/n) x log Q

Where:

n: number of electrons that are transferred in the reaction. In this case, n= 2.

Q: ratio between the concentrations of products over reactants, calculated as follows:

Q = \frac{ [Co^{2+} ]}{[Fe^{2+} ]} = \frac{0.33 M}{1.47 M} = 0.2244

Finally, we introduce Eº= 0.167 V, n= 2, Q=0.2244, to obtain E:

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How many grams of vanadium would contain the same number of atoms as 58.693 g of nickel
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To answer this question, we'll need the molar masses of nickel and vanadium:

Molar mass of nickel: 58.693g/mol

Molar mass of vanadium: 50.941g/mol

These values represent the mass of one mole of the element. A mole is the weight of Avogadro's number of atoms of an element (6.02 x 10^23 atoms).

We are given 58.693g of nickel, which is equivalent to one mole of nickel. Therefore, you will get the same number of atoms from one mole of vanadium, or 50.941g of vanadium.

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